新高考數(shù)學(xué)一輪復(fù)習(xí)第2章 第07講 函數(shù)的圖象精講+精練（教師版）

第07講函數(shù)的圖象(精講+精練）目錄第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第二部分：課前自我評(píng)估測(cè)試第三部分：典型例題剖析高頻考點(diǎn)一：畫出函數(shù)的圖象高頻考點(diǎn)二：函數(shù)圖象的識(shí)別高頻考點(diǎn)三：函數(shù)圖象的應(yīng)用①研究函數(shù)的性質(zhì)②確定零點(diǎn)個(gè)數(shù)③解不等式④求參數(shù)的取值范圍第四部分：高考真題感悟第五部分：第07講函數(shù)的圖象（精練）第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶第一部分：知識(shí)點(diǎn)精準(zhǔn)記憶1、平移變換（左“+”右“-”；上“+”下“-”）①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0④SKIPIF1<0注：左右平移只能單獨(dú)一個(gè)SKIPIF1<0加或者減，注意當(dāng)SKIPIF1<0前系數(shù)不為1,需將系數(shù)提取到外面.2、對(duì)稱變換①SKIPIF1<0的圖象SKIPIF1<0SKIPIF1<0的圖象；②SKIPIF1<0的圖象SKIPIF1<0SKIPIF1<0的圖象；③SKIPIF1<0的圖象SKIPIF1<0SKIPIF1<0的圖象；④SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)的圖象SKIPIF1<0SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)的圖象.3、伸縮變換①SKIPIF1<0SKIPIF1<0SKIPIF1<0.②SKIPIF1<0SKIPIF1<0SKIPIF1<0.4、翻折變換（絕對(duì)值變換）①SKIPIF1<0的圖象SKIPIF1<0SKIPIF1<0的圖象；（口訣；以SKIPIF1<0軸為界，保留SKIPIF1<0軸上方的圖象；將SKIPIF1<0軸下方的圖象翻折到SKIPIF1<0軸上方）②SKIPIF1<0的圖象SKIPIF1<0SKIPIF1<0的圖象.（口訣；以SKIPIF1<0軸為界，去掉SKIPIF1<0軸左側(cè)的圖象，保留SKIPIF1<0軸右側(cè)的圖象；將SKIPIF1<0軸右側(cè)圖象翻折到SKIPIF1<0軸左側(cè)；本質(zhì)是個(gè)偶函數(shù)）5、圖象識(shí)別技巧（按使用頻率優(yōu)先級(jí)排序）①特殊值法（觀察圖象，尋找圖象中出現(xiàn)的特殊值）②單調(diào)性法（SKIPIF1<0；SKIPIF1<0；SKIPIF1<0，SKIPIF1<0；通過(guò)求導(dǎo)判斷單調(diào)性）③奇偶性法SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0偶函數(shù)偶函數(shù)偶函數(shù)偶函數(shù)偶函數(shù)偶函數(shù)偶函數(shù)奇函數(shù)不能確定不能確定奇函數(shù)奇函數(shù)奇函數(shù)偶函數(shù)不能確定不能確定奇函數(shù)奇函數(shù)奇函數(shù)奇函數(shù)奇函數(shù)奇函數(shù)偶函數(shù)偶函數(shù)④極限（左右極限）（SKIPIF1<0；SKIPIF1<0；SKIPIF1<0；SKIPIF1<0；）⑤零點(diǎn)法⑥極大值極小值法第二部分：課前自我評(píng)估測(cè)試第二部分：課前自我評(píng)估測(cè)試1．（2022·陜西西安·高一期末）函數(shù)SKIPIF1<0的圖像大致為（

）A． B．C． D．【答案】CSKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，圖象關(guān)于原點(diǎn)對(duì)稱，所以AD選項(xiàng)錯(cuò)誤.SKIPIF1<0，所以B選項(xiàng)錯(cuò)誤.故選：C2．（2022·北京·高三學(xué)業(yè)考試）函數(shù)SKIPIF1<0的圖象如圖所示，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C由圖象可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.故選：C3．（2022·全國(guó)·高三專題練習(xí)）在同一直角坐標(biāo)系中，函數(shù)SKIPIF1<0，SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】DA：沒(méi)有冪函數(shù)圖象，不符合；B：SKIPIF1<0中SKIPIF1<0，SKIPIF1<0中SKIPIF1<0，不符合；C：SKIPIF1<0中SKIPIF1<0，SKIPIF1<0中SKIPIF1<0，不符合；D：SKIPIF1<0中SKIPIF1<0，SKIPIF1<0中SKIPIF1<0，符合.故選：D.4．（2022·浙江金華第一中學(xué)高一期末）圖（1）是某條公共汽車線路收支差額SKIPIF1<0關(guān)于乘客量SKIPIF1<0的圖象，圖（2）?（3）是由于目前本條路線虧損，公司有關(guān)人員提出的兩種扭虧為盈的建議，則下列說(shuō)法錯(cuò)誤的是（

）A．圖（1）的點(diǎn)SKIPIF1<0的實(shí)際意義為：當(dāng)乘客量為0時(shí)，虧損1個(gè)單位B．圖（1）的射線SKIPIF1<0上的點(diǎn)表示當(dāng)乘客量小于3時(shí)將虧損，大于3時(shí)將盈利C．圖（2）的建議為降低成本而保持票價(jià)不變D．圖（3）的建議為降低成本的同時(shí)提高票價(jià)【答案】DA：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)乘客量為0時(shí)，虧損1個(gè)單位，故本選項(xiàng)說(shuō)法正確；B：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以本選項(xiàng)說(shuō)法正確；C：降低成本而保持票價(jià)不變，兩條線是平行，所以本選項(xiàng)正確；D：由圖可知中：成本不變，同時(shí)提高票價(jià)，所以本選項(xiàng)說(shuō)法不正確，故選：D第三部分：典型例題剖析第三部分：典型例題剖析高頻考點(diǎn)一：畫出函數(shù)的圖象1．（2021·寧夏·銀川市第六中學(xué)高一期中）已知函數(shù)SKIPIF1<0.(1)證明SKIPIF1<0是偶函數(shù)；(2)畫出這個(gè)函數(shù)的圖象；(3)求函數(shù)SKIPIF1<0的值域.【答案】(1)證明見解析；(2)圖象見解析；(3)SKIPIF1<0(1)解：由題知函數(shù)的定義域關(guān)于原點(diǎn)對(duì)稱，SKIPIF1<0，所以函數(shù)SKIPIF1<0是偶函數(shù)SKIPIF1<0(2)解：由題知SKIPIF1<0，進(jìn)而結(jié)合二次函數(shù)與分段函數(shù)的性質(zhì)作圖如下：(3)解：由（2）的函數(shù)圖象可知函數(shù)的最小值為SKIPIF1<0，函數(shù)的最大值為SKIPIF1<0，所以函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<02．（2021·山東臨沂·高一期中）已知SKIPIF1<0是整數(shù)，冪函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．(1)求SKIPIF1<0的解析式；(2)若SKIPIF1<0，畫出函數(shù)SKIPIF1<0的大致圖象；(3)寫出SKIPIF1<0的單調(diào)區(qū)間，并用定義法證明SKIPIF1<0在區(qū)間SKIPIF1<0上的單調(diào)性．【答案】(1)SKIPIF1<0(2)作圖見解析(3)單調(diào)減區(qū)間為SKIPIF1<0，SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0，SKIPIF1<0，證明見解析(1)解：由題意可知，SKIPIF1<0，即SKIPIF1<0，因?yàn)镾KIPIF1<0是整數(shù)，所以SKIPIF1<0，或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，綜上可知，SKIPIF1<0的解析式為SKIPIF1<0；(2)解：由（1）知SKIPIF1<0，則SKIPIF1<0，函數(shù)SKIPIF1<0的圖象如圖所示，(3)解：由（2）可知，SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0，SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，設(shè)任意的SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0，∵SKIPIF1<0，且SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增．3．（2021·全國(guó)·高一課時(shí)練習(xí)）根據(jù)SKIPIF1<0的圖像，作出下列函數(shù)的圖像：(1)SKIPIF1<0；(2)SKIPIF1<0；(3)SKIPIF1<0；(4)SKIPIF1<0．(1)作出函數(shù)SKIPIF1<0關(guān)于縱軸對(duì)稱的圖像，連同函數(shù)SKIPIF1<0的圖像，就是該函數(shù)的圖像，如下圖所示：(2)把函數(shù)SKIPIF1<0的圖像中縱軸下面的部分，做關(guān)于橫軸對(duì)稱，擦掉縱軸下面的部分，函數(shù)圖像如下圖所示：(3)作出函數(shù)SKIPIF1<0關(guān)于縱軸對(duì)稱的圖像，連同函數(shù)SKIPIF1<0的圖像一起向右平移一個(gè)單位即可，如下圖所示：(4)把函數(shù)SKIPIF1<0的圖像中縱軸下面的部分，做關(guān)于橫軸對(duì)稱，擦掉縱軸下面的部分，然后再向右平移一個(gè)單位，如下圖所示：高頻考點(diǎn)二：函數(shù)圖象的識(shí)別1．（2022·福建福州·高一期末）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的大致圖像為（

）A． B．C． D．【答案】B解：由題得SKIPIF1<0，所以排除選項(xiàng)A,D.SKIPIF1<0,所以排除選項(xiàng)C.故選：B2．（2022·湖南·株洲二中高一階段練習(xí)）函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】D由題意知，SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0定義域SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱，又因?yàn)镾KIPIF1<0，所以此函數(shù)為奇函數(shù)，圖像關(guān)于原點(diǎn)對(duì)稱，排除A.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除B.SKIPIF1<0，函數(shù)只有1個(gè)零點(diǎn)，排除C.故選：D3．（2022·山東德州·高三期末）已知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的大致圖象為（

）A． B．C． D．【答案】D由題可知：函數(shù)定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故該函數(shù)為奇函數(shù)，排除A,C又SKIPIF1<0，所以排除B,故選：D4．（2022·浙江·高三學(xué)業(yè)考試）函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】B因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镽，且SKIPIF1<0不是偶函數(shù)，所以排除C、D；又SKIPIF1<0，排除A，即確定答案為B.故選:B.5．（2022·全國(guó)·高三專題練習(xí)（文））函數(shù)SKIPIF1<0的部分圖象大致是（

）A． B．C． D．【答案】A由題可知函數(shù)定義域?yàn)镾KIPIF1<0，則SKIPIF1<0，又SKIPIF1<0所以SKIPIF1<0是奇函數(shù)，且SKIPIF1<0時(shí)，SKIPIF1<0，故選項(xiàng)A正確.

故選：A6．（2022·廣西南寧·一模（文））函數(shù)SKIPIF1<0的圖象最有可能是以下的（

）A． B．C． D．【答案】BSKIPIF1<0定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱，又SKIPIF1<0，所以SKIPIF1<0是奇函數(shù)，故排除CD，又SKIPIF1<0，故排除A選項(xiàng)，B正確.故選：B7．（2022·陜西咸陽(yáng)·高一期末）函數(shù)SKIPIF1<0的大致圖像為（

）A． B．C． D．【答案】D對(duì)任意的SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，排除C選項(xiàng)；SKIPIF1<0，SKIPIF1<0，所以，函數(shù)SKIPIF1<0為偶函數(shù)，排除B選項(xiàng)，因?yàn)镾KIPIF1<0，排除A選項(xiàng).故選：D.高頻考點(diǎn)三：函數(shù)圖象的應(yīng)用①研究函數(shù)的性質(zhì)1．（2022·河南·林州一中高一開學(xué)考試）已知函數(shù)SKIPIF1<0，SKIPIF1<0則（

）A．SKIPIF1<0的最大值為3，最小值為1B．SKIPIF1<0的最大值為SKIPIF1<0，無(wú)最小值C．SKIPIF1<0的最大值為SKIPIF1<0，最小值為1D．SKIPIF1<0的最大值為3，最小值為-1【答案】B解：SKIPIF1<0，由SKIPIF1<0與SKIPIF1<0，解SKIPIF1<0得SKIPIF1<0；解SKIPIF1<0得SKIPIF1<0；所以SKIPIF1<0與SKIPIF1<0的交點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的圖象如下圖所示：由圖象，可知最大值為SKIPIF1<0，無(wú)最小值，故選：B．2．（2022·全國(guó)·高一期末）已知函數(shù)SKIPIF1<0.(1)判斷函數(shù)SKIPIF1<0的奇偶性，并證明；(2)畫出函數(shù)SKIPIF1<0的圖象，并討論方程SKIPIF1<0的解的個(gè)數(shù).【答案】(1)函數(shù)SKIPIF1<0為偶函數(shù)，證明見解析；(2)圖象見解析；當(dāng)SKIPIF1<0時(shí)，方程的解為0個(gè)；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，方程的解為2個(gè)；當(dāng)SKIPIF1<0時(shí)，方程的解為4個(gè)；當(dāng)SKIPIF1<0時(shí)，方程的解為3個(gè).(1)函數(shù)SKIPIF1<0為偶函數(shù)，∵SKIPIF1<0，∴SKIPIF1<0的定義域?yàn)镽，關(guān)于原點(diǎn)對(duì)稱，且SKIPIF1<0，所以SKIPIF1<0為偶函數(shù)；(2)因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0的圖象如下所示：方程SKIPIF1<0的解的個(gè)數(shù)，即SKIPIF1<0與SKIPIF1<0的交點(diǎn)個(gè)數(shù)，結(jié)合函數(shù)圖象可知：當(dāng)SKIPIF1<0時(shí)，有0個(gè)解，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，有2個(gè)解，當(dāng)SKIPIF1<0時(shí)，有4個(gè)解，當(dāng)SKIPIF1<0時(shí)，有3個(gè)解.3．（2022·山東濰坊·高一期末）已知定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.(1)求函數(shù)SKIPIF1<0在SKIPIF1<0上的解析式；(2)在給出的直角坐標(biāo)系中作出SKIPIF1<0的圖像，并寫出函數(shù)SKIPIF1<0的單調(diào)區(qū)間.【答案】(1)SKIPIF1<0(2)圖像答案見解析，單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0(1)設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0，又SKIPIF1<0為定義在SKIPIF1<0上的奇函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0(2)作出函數(shù)SKIPIF1<0的圖像，如圖所示：函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.②確定零點(diǎn)個(gè)數(shù)1．（2022·全國(guó)·高三階段練習(xí)）函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D令SKIPIF1<0，得SKIPIF1<0；在同一直角坐標(biāo)系中分別作出SKIPIF1<0，SKIPIF1<0的大致圖象如圖所示；觀察可知，兩個(gè)函數(shù)的圖象有SKIPIF1<0個(gè)交點(diǎn)（其中SKIPIF1<0個(gè)交點(diǎn)的橫坐標(biāo)介于SKIPIF1<0到SKIPIF1<0之間，另外兩個(gè)交點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，故函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)為SKIPIF1<0，故選：D．2．（2022·江西·高一期末）已知函數(shù)SKIPIF1<0，若方程SKIPIF1<0恰有兩個(gè)不等的實(shí)根，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B方程SKIPIF1<0恰有兩個(gè)不等的實(shí)根，等價(jià)于SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，SKIPIF1<0的圖象如圖所示，平移水平直線SKIPIF1<0可得SKIPIF1<0，故選：B.3．（2022·湖南·長(zhǎng)沙市南雅中學(xué)高三階段練習(xí)）函數(shù)SKIPIF1<0有三個(gè)不同的零點(diǎn)，則實(shí)數(shù)t的范圍是__________.【答案】SKIPIF1<0作出函數(shù)SKIPIF1<0的圖象和直線SKIPIF1<0，如圖，由圖象可得SKIPIF1<0時(shí)，直線與函數(shù)圖象有三個(gè)交點(diǎn)，即函數(shù)SKIPIF1<0有三個(gè)零點(diǎn)．SKIPIF1<0．故答案為：SKIPIF1<0．4．（2022·湖南·高一課時(shí)練習(xí)）用圖象法判定方程SKIPIF1<0的根的個(gè)數(shù).【答案】1方程SKIPIF1<0根的個(gè)數(shù)，等價(jià)于函數(shù)SKIPIF1<0與SKIPIF1<0圖象的交點(diǎn)個(gè)數(shù)，函數(shù)SKIPIF1<0與SKIPIF1<0在同一坐標(biāo)系中的圖象如圖所示，兩函數(shù)圖象只有一個(gè)交點(diǎn)，所以方程SKIPIF1<0的根的個(gè)數(shù)為1③解不等式1．（2022·全國(guó)·池州市第一中學(xué)高一開學(xué)考試）已知函數(shù)SKIPIF1<0的圖象如圖，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D不等式SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，觀察圖象，解SKIPIF1<0得SKIPIF1<0，解SKIPIF1<0得SKIPIF1<0，所以不等式SKIPIF1<0的解集為SKIPIF1<0.故選：D2．（2022·河北·高三階段練習(xí)）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象，如圖，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象，由圖象可知，此時(shí)解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象，它們的交點(diǎn)坐標(biāo)為SKIPIF1<0、SKIPIF1<0，結(jié)合圖象知此時(shí)SKIPIF1<0.所以不等式SKIPIF1<0的解集為SKIPIF1<0SKIPIF1<0.故選：C3．（2022·北京·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B不等式SKIPIF1<0，分別畫出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象，由圖象可知SKIPIF1<0和SKIPIF1<0有兩個(gè)交點(diǎn)，分別是SKIPIF1<0和SKIPIF1<0，由圖象可知SKIPIF1<0的解集是SKIPIF1<0即不等式SKIPIF1<0的解集是SKIPIF1<0.故選：B4．（2022·河南·信陽(yáng)高中高一階段練習(xí)（理））已知定義在R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若對(duì)任SKIPIF1<0都有SKIPIF1<0，則m的取值范圍是_________．【答案】SKIPIF1<0，SKIPIF1<0．解：因?yàn)镾KIPIF1<0滿足SKIPIF1<0，即SKIPIF1<0；又由SKIPIF1<0，可得SKIPIF1<0，畫出當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0的圖象，將SKIPIF1<0在SKIPIF1<0，SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位（橫坐標(biāo)不變，縱坐標(biāo)變?yōu)樵瓉?lái)的2倍），再向左平移SKIPIF1<0個(gè)單位（橫坐標(biāo)不變，縱坐標(biāo)變?yōu)樵瓉?lái)的SKIPIF1<0倍），由此得到函數(shù)SKIPIF1<0的圖象如圖：當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，由圖像可得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，滿足對(duì)任意的SKIPIF1<0，SKIPIF1<0，都有SKIPIF1<0，故SKIPIF1<0的范圍為SKIPIF1<0，SKIPIF1<0．故答案為：SKIPIF1<0，SKIPIF1<0．④求參數(shù)的取值范圍1．（2022·山西·靈丘縣第一中學(xué)校高二階段練習(xí)）已知函數(shù)SKIPIF1<0若直線SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C設(shè)SKIPIF1<0與SKIPIF1<0相切于點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，此時(shí)SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，此時(shí)切點(diǎn)為SKIPIF1<0，作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象如圖，由圖象可知，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0有三個(gè)不同的交點(diǎn)，故選：C2．（2022·河北石家莊·高一期末）已知函數(shù)SKIPIF1<0，若存在不相等的實(shí)數(shù)a，b，c，d滿足SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C由題設(shè)，將問(wèn)題轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0的圖象有四個(gè)交點(diǎn)，SKIPIF1<0，則在SKIPIF1<0上遞減且值域?yàn)镾KIPIF1<0；在SKIPIF1<0上遞增且值域?yàn)镾KIPIF1<0；在SKIPIF1<0上遞減且值域?yàn)镾KIPIF1<0，在SKIPIF1<0上遞增且值域?yàn)镾KIPIF1<0；SKIPIF1<0的圖象如下：所以SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0的圖象有四個(gè)交點(diǎn)，不妨假設(shè)SKIPIF1<0，由圖及函數(shù)性質(zhì)知：SKIPIF1<0，易知：SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故選：C3．（多選）（2022·湖北·石首市第一中學(xué)高一階段練習(xí)）函數(shù)SKIPIF1<0恰有2個(gè)零點(diǎn)，則SKIPIF1<0的取值可以是（

）A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】BD解：由題意得：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，該函數(shù)是由SKIPIF1<0向上或向下平移SKIPIF1<0個(gè)單位得到當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0對(duì)于函數(shù)SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0若SKIPIF1<0，即SKIPIF1<0，函數(shù)SKIPIF1<0與SKIPIF1<0軸沒(méi)有交點(diǎn)，則SKIPIF1<0滿足不等式組SKIPIF1<0故可取SKIPIF1<0，如圖1所示；若SKIPIF1<0，即SKIPIF1<0，函數(shù)SKIPIF1<0與SKIPIF1<0軸有一個(gè)交點(diǎn)，則SKIPIF1<0滿足不等式SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0或無(wú)解，如圖2所示；又SKIPIF1<0，解得SKIPIF1<0，故可取SKIPIF1<0故選：BD4．（2022·河南·欒川縣第一高級(jí)中學(xué)高二階段練習(xí)（理））已知函數(shù)SKIPIF1<0若SKIPIF1<0且SKIPIF1<0，則SKIPIF1<0的最小值是________．【答案】SKIPIF1<0##SKIPIF1<0函數(shù)SKIPIF1<0的圖象如圖所示．令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0．令SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0．故答案為：SKIPIF1<0．5．（2022·山西臨汾·二模（理））已知函數(shù)SKIPIF1<0有2個(gè)不同的零點(diǎn)，則k的取值范圍是____________．【答案】SKIPIF1<0因?yàn)楹瘮?shù)SKIPIF1<0有2個(gè)不同的零點(diǎn)，所以關(guān)于SKIPIF1<0的方程SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有兩個(gè)不等的實(shí)根，即曲線SKIPIF1<0（圓SKIPIF1<0的上半部分）與經(jīng)過(guò)定點(diǎn)SKIPIF1<0的直線SKIPIF1<0有兩個(gè)不同的交點(diǎn)，如圖過(guò)SKIPIF1<0作圓SKIPIF1<0的切線SKIPIF1<0，則點(diǎn)SKIPIF1<0到切線SKIPIF1<0的距離SKIPIF1<0，解得SKIPIF1<0（舍去）或SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，即k的取值范圍是SKIPIF1<0，故答案為：SKIPIF1<0第四部分：高考真題感悟第四部分：高考真題感悟1．（2021·天津·高考真題）函數(shù)SKIPIF1<0的圖像大致為（

）A． B．C． D．【答案】B設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱，又SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，排除AC；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除D.故選：B.2．（2021·浙江·高考真題）已知函數(shù)SKIPIF1<0，則圖象為如圖的函數(shù)可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D對(duì)于A，SKIPIF1<0，該函數(shù)為非奇非偶函數(shù)，與函數(shù)圖象不符，排除A；對(duì)于B，SKIPIF1<0，該函數(shù)為非奇非偶函數(shù)，與函數(shù)圖象不符，排除B；對(duì)于C，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，與圖象不符，排除C.故選：D.3．（2020·天津·高考真題）函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】A由函數(shù)的解析式可得：SKIPIF1<0，則函數(shù)SKIPIF1<0為奇函數(shù)，其圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，選項(xiàng)CD錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，選項(xiàng)B錯(cuò)誤.故選：A.【點(diǎn)睛】函數(shù)圖象的識(shí)辨可從以下方面入手：(1)從函數(shù)的定義域，判斷圖象的左右位置；從函數(shù)的值域，判斷圖象的上下位置．(2)從函數(shù)的單調(diào)性，判斷圖象的變化趨勢(shì)．(3)從函數(shù)的奇偶性，判斷圖象的對(duì)稱性．(4)從函數(shù)的特征點(diǎn)，排除不合要求的圖象．利用上述方法排除、篩選選項(xiàng)．4．（2020·北京·高考真題）已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集是（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D因?yàn)镾KIPIF1<0，所以SKIPIF1<0等價(jià)于SKIPIF1<0，在同一直角坐標(biāo)系中作出SKIPIF1<0和SKIPIF1<0的圖象如圖：兩函數(shù)圖象的交點(diǎn)坐標(biāo)為SKIPIF1<0，不等式SKIPIF1<0的解為SKIPIF1<0或SKIPIF1<0.所以不等式SKIPIF1<0的解集為：SKIPIF1<0.故選：D.5．（2020·浙江·高考真題）函數(shù)y=xcosx+sinx在區(qū)間[–π，π]的圖象大致為（）A． B．C． D．【答案】A【詳解】因?yàn)镾KIPIF1<0，則SKIPIF1<0，即題中所給的函數(shù)為奇函數(shù)，函數(shù)圖象關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，據(jù)此可知選項(xiàng)CD錯(cuò)誤；且SKIPIF1<0時(shí)，SKIPIF1<0，據(jù)此可知選項(xiàng)B錯(cuò)誤.故選：A.【點(diǎn)睛】函數(shù)圖象的識(shí)辨可從以下方面入手：(1)從函數(shù)的定義域，判斷圖象的左右位置；從函數(shù)的值域，判斷圖象的上下位置．(2)從函數(shù)的單調(diào)性，判斷圖象的變化趨勢(shì)．(3)從函數(shù)的奇偶性，判斷圖象的對(duì)稱性．(4)從函數(shù)的特征點(diǎn)，排除不合要求的圖象．利用上述方法排除、篩選選項(xiàng)．6．（2021·湖南·高考真題）已知函數(shù)SKIPIF1<0（1）畫出函數(shù)SKIPIF1<0的圖象；（2）若SKIPIF1<0，求SKIPIF1<0的取值范圍.【答案】（1）答案見解析；（2）SKIPIF1<0（1）函數(shù)SKIPIF1<0的圖象如圖所示：（2）SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，可得：SKIPIF1<0，當(dāng)SKIPIF1<0，SKIPIF1<0，可得：SKIPIF1<0，所以SKIPIF1<0的解集為：SKIPIF1<0，所以SKIPIF1<0的取值范圍為SKIPIF1<0.第五部分：第07講函數(shù)的圖象（精練）第五部分：第07講函數(shù)的圖象（精練）一、單選題1．（2022·湖南·高一課時(shí)練習(xí)）函數(shù)SKIPIF1<0與SKIPIF1<0的定義域均為SKIPIF1<0，它們的圖象如圖所示，則不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A解：SKIPIF1<0即為函數(shù)SKIPIF1<0的圖像在函數(shù)SKIPIF1<0的圖像的上方的部分對(duì)應(yīng)自變量的范圍，由圖可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，即不等式SKIPIF1<0的解集是SKIPIF1<0.故選：A.2．（2022·全國(guó)·高三專題練習(xí)）如圖為函數(shù)SKIPIF1<0的圖象，則該函數(shù)可能為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B由圖可知，SKIPIF1<0時(shí)，SKIPIF1<0，ACD的函數(shù)SKIPIF1<0,故選：B.3．（2022·北京交通大學(xué)附屬中學(xué)高二階段練習(xí)）函數(shù)SKIPIF1<0的圖象大致為（

）A．B．C． D．【答案】C根據(jù)題意，函數(shù)SKIPIF1<0，其定義域?yàn)镾KIPIF1<0，有SKIPIF1<0，即函數(shù)SKIPIF1<0為奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，函數(shù)的圖象在第一象限，分析選項(xiàng)可得：C符合故選：C【點(diǎn)睛】思路點(diǎn)睛：函數(shù)圖象的辨識(shí)可從以下方面入手：(1)從函數(shù)的定義域，判斷圖象的左右位置；從函數(shù)的值域，判斷圖象的上下位置．(2)從函數(shù)的單調(diào)性，判斷圖象的變化趨勢(shì)；(3)從函數(shù)的奇偶性，判斷圖象的對(duì)稱性；(4)從函數(shù)的特征點(diǎn)，排除不合要求的圖象.4．（2022·浙江紹興·模擬預(yù)測(cè)）在同一直角坐標(biāo)系中，函數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0的圖象可能是（

）A． B．C． D．【答案】C解：因?yàn)楹瘮?shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，所以函數(shù)SKIPIF1<0的圖象恒過(guò)定點(diǎn)SKIPIF1<0，故選項(xiàng)A、B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，又SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，故選項(xiàng)D錯(cuò)誤，選項(xiàng)C正確.故選：C.5．（2022·新疆·模擬預(yù)測(cè)（理））我國(guó)著名數(shù)學(xué)家華羅庚曾說(shuō)：“數(shù)缺形時(shí)少直觀，形缺數(shù)時(shí)難入微，數(shù)形結(jié)合百般好，隔裂分家萬(wàn)事休.”在數(shù)學(xué)的學(xué)習(xí)和研究中，常用函數(shù)的圖象來(lái)研究函數(shù)的性質(zhì)，也常用函數(shù)的解析式來(lái)研究函數(shù)圖象的特征.我們從這個(gè)商標(biāo)中抽象出一個(gè)函數(shù)的圖象如圖，其對(duì)應(yīng)的函數(shù)解析式可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】DA：函數(shù)的定義域?yàn)镾KIPIF1<0，不符合；B：由SKIPIF1<0，不符合；C：由SKIPIF1<0，不符合；D：SKIPIF1<0且定義域?yàn)镾KIPIF1<0，SKIPIF1<0為偶函數(shù)，在SKIPIF1<0上SKIPIF1<0單調(diào)遞增，SKIPIF1<0上SKIPIF1<0單調(diào)遞減，結(jié)合偶函數(shù)的對(duì)稱性知：SKIPIF1<0上遞減，SKIPIF1<0上遞增，符合.故選：D6．（2022·四川達(dá)州·二模（理））函數(shù)SKIPIF1<0的部分圖象大致為（

）A． B．C． D．【答案】A∵函數(shù)SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，故排除BD；又SKIPIF1<0，故排除C.故選：A.7．（2022·全國(guó)·高三專題練習(xí)（理））已知函數(shù)SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0均不相等，且SKIPIF1<0=SKIPIF1<0=SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．（1，10） B．(5，6) C．(10，12) D．(20，24)【答案】C畫出函數(shù)的圖象，如圖所示，不妨設(shè)SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0，SKIPIF1<0的取值范圍是SKIPIF1<0，所以SKIPIF1<0的取值范圍是SKIPIF1<0．故選：C8．（2022·貴州畢節(jié)·模擬預(yù)測(cè)（理））設(shè)SKIPIF1<0是定義在R上且周期為2的函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其中a，SKIPIF1<0，且函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，則a的取值不可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．0【答案】D因?yàn)镾KIPIF1<0是定義在R上且周期為2的函數(shù)，所以SKIPIF1<0，所以SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，其圖象如圖所示SKIPIF1<0，由于周期為2，所以SKIPIF1<0，所以SKIPIF1<0不符合題意，當(dāng)SKIPIF1<0時(shí)，則圖象向上平移，函數(shù)無(wú)零點(diǎn)，所以不符合，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0在SKIPIF1<0上有一個(gè)零點(diǎn)，所以SKIPIF1<0在SKIPIF1<0上有零點(diǎn)，所以SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，符合題意，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0在SKIPIF1<0上有2個(gè)零點(diǎn)，由于函數(shù)的周期為2，所以SKIPIF1<0在SKIPIF1<0上有6個(gè)零點(diǎn)，不符合題意，當(dāng)SKIPIF1<0時(shí)，則可得SKIPIF1<0，在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，所以符合題意，當(dāng)SKIPIF1<0時(shí)，函數(shù)圖象與SKIPIF1<0軸無(wú)交點(diǎn)，綜上，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0在區(qū)間SKIPIF1<0上恰有3個(gè)零點(diǎn)，故選：D二、填空題9．（2022·重慶·高一期末）已知冪函數(shù)SKIPIF1<0的圖象如圖所示，則SKIPIF1<0______．（寫出一個(gè)正確結(jié)果即可）【答案】SKIPIF1<0（答案不唯一）由冪函數(shù)圖象知，函數(shù)SKIPIF1<0的定義域是SKIPIF1<0，且在SKIPIF1<0單調(diào)遞減，于是得冪函數(shù)的冪指數(shù)為負(fù)數(shù)，而函數(shù)SKIPIF1<0的圖象關(guān)于y軸對(duì)稱，即冪函數(shù)SKIPIF1<0是偶函數(shù)，則冪函數(shù)SKIPIF1<0的冪指數(shù)為偶數(shù)，綜上得：SKIPIF1<0.故答案為：SKIPIF1<010．（2022·山東威海·高一期末）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有四個(gè)根，則實(shí)數(shù)SKIPIF1<0的取值范圍為______．【答案】SKIPIF1<0由SKIPIF1<0，得SKIPIF1<0令SKIPIF1<0，畫出圖像由圖可知，當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有四解，即方程SKIPIF1<0有四個(gè)根.故答案為:SKIPIF1<011．（2022·云南·高三階段練習(xí)（理））函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0恰有4個(gè)零點(diǎn)，則實(shí)數(shù)m的取值范圍是_________．【答案】SKIPIF1<0由題意，函數(shù)SKIPIF1<0恰有4個(gè)零點(diǎn)，即SKIPIF1<0，即SKIPIF1<0有4個(gè)不同的實(shí)數(shù)根，即直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有四個(gè)不同的交點(diǎn)，又由SKIPIF1<0，作出該函數(shù)SKIPIF1<0的圖象，如圖所示，當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，其中SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0，其中SKIPIF1<0時(shí)，SKIPIF1<0，結(jié)合圖象可得，當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有4個(gè)不同的交點(diǎn)，即函數(shù)SKIPIF1<0恰有4個(gè)零點(diǎn)時(shí)，所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.12．（2022·重慶·高一期末）設(shè)SKIPIF1<0函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的方程SKIPIF1<0有三個(gè)不相等的實(shí)數(shù)解，則實(shí)數(shù)SKIPIF1<0的取值范圍是______．【答案】